Package g1601_1700.s1629_slowest_key
Class Solution
java.lang.Object
g1601_1700.s1629_slowest_key.Solution
1629 - Slowest Key.<p>Easy</p>
<p>A newly designed keypad was tested, where a tester pressed a sequence of <code>n</code> keys, one at a time.</p>
<p>You are given a string <code>keysPressed</code> of length <code>n</code>, where <code>keysPressed[i]</code> was the <code>i<sup>th</sup></code> key pressed in the testing sequence, and a sorted list <code>releaseTimes</code>, where <code>releaseTimes[i]</code> was the time the <code>i<sup>th</sup></code> key was released. Both arrays are <strong>0-indexed</strong>. The <code>0<sup>th</sup></code> key was pressed at the time <code>0</code>, and every subsequent key was pressed at the <strong>exact</strong> time the previous key was released.</p>
<p>The tester wants to know the key of the keypress that had the <strong>longest duration</strong>. The <code>i<sup>th</sup></code> keypress had a <strong>duration</strong> of <code>releaseTimes[i] - releaseTimes[i - 1]</code>, and the <code>0<sup>th</sup></code> keypress had a duration of <code>releaseTimes[0]</code>.</p>
<p>Note that the same key could have been pressed multiple times during the test, and these multiple presses of the same key <strong>may not</strong> have had the same <strong>duration</strong>.</p>
<p><em>Return the key of the keypress that had the <strong>longest duration</strong>. If there are multiple such keypresses, return the lexicographically largest key of the keypresses.</em></p>
<p><strong>Example 1:</strong></p>
<p><strong>Input:</strong> releaseTimes = [9,29,49,50], keysPressed = “cbcd”</p>
<p><strong>Output:</strong> “c”</p>
<p><strong>Explanation:</strong> The keypresses were as follows:</p>
<p>Keypress for ‘c’ had a duration of 9 (pressed at time 0 and released at time 9).</p>
<p>Keypress for ‘b’ had a duration of 29 - 9 = 20 (pressed at time 9 right after the release of the previous character and released at time 29).</p>
<p>Keypress for ‘c’ had a duration of 49 - 29 = 20 (pressed at time 29 right after the release of the previous character and released at time 49).</p>
<p>Keypress for ‘d’ had a duration of 50 - 49 = 1 (pressed at time 49 right after the release of the previous character and released at time 50).</p>
<p>The longest of these was the keypress for ‘b’ and the second keypress for ‘c’, both with duration 20. ‘c’ is lexicographically larger than ‘b’, so the answer is ‘c’.</p>
<p><strong>Example 2:</strong></p>
<p><strong>Input:</strong> releaseTimes = [12,23,36,46,62], keysPressed = “spuda”</p>
<p><strong>Output:</strong> “a”</p>
<p><strong>Explanation:</strong></p>
<p>The keypresses were as follows: Keypress for ‘s’ had a duration of 12.</p>
<p>Keypress for ‘p’ had a duration of 23 - 12 = 11.</p>
<p>Keypress for ‘u’ had a duration of 36 - 23 = 13.</p>
<p>Keypress for ‘d’ had a duration of 46 - 36 = 10.</p>
<p>Keypress for ‘a’ had a duration of 62 - 46 = 16.</p>
<p>The longest of these was the keypress for ‘a’ with duration 16.</p>
<p><strong>Constraints:</strong></p>
<ul>
<li><code>releaseTimes.length == n</code></li>
<li><code>keysPressed.length == n</code></li>
<li><code>2 <= n <= 1000</code></li>
<li><code>1 <= releaseTimes[i] <= 10<sup>9</sup></code></li>
<li><code>releaseTimes[i] < releaseTimes[i+1]</code></li>
<li><code>keysPressed</code> contains only lowercase English letters.</li>
</ul>
-
Constructor Summary
Constructors -
Method Summary
-
Constructor Details
-
Solution
public Solution()
-
-
Method Details
-
slowestKey
-